Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $20$ years; the standard deviation is $3.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than $12.4$ years.
Solution: $20$ $16.2$ $23.8$ $12.4$ $27.6$ $8.6$ $31.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $20$ years. We know the standard deviation is $3.8$ years, so one standard deviation below the mean is $16.2$ years and one standard deviation above the mean is $23.8$ years. Two standard deviations below the mean is $12.4$ years and two standard deviations above the mean is $27.6$ years. Three standard deviations below the mean is $8.6$ years and three standard deviations above the mean is $31.4$ years. We are interested in the probability of a zebra living less than $12.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $12.4$ years and the other half $({2.5\%})$ will live longer than $27.6$ years. The probability of a particular zebra living less than $12.4$ years is ${2.5\%}$.